How many times does cos(96πt)cos(4πt)=0 during t=0 to t=1s?

[zorro]

Homework Statement

This problem came up while solving a physics problem in waves.
We have the equation cos(96πt)cos(4πt)=0
How many times does the L.H.S. become 0 during the time t=0 to t=1s ?

The Attempt at a Solution

Nothing.

 

[HallsofIvy]

A product of numbers is 0 only if at least one number is 0. cos(x) is 0 when x is an odd multiple of \pi/2. For what values of t is 192t an odd integer? For what values of t is 8t an odd integer?

 

[zorro]

For what values of t is 192t an odd integer?

t can be 1/192, 1/64 and for second case t=1/8 only. But 3 times is not the correct answer.

 

[HallsofIvy]

I have no idea what you are doing! I get 192/2= 96 values of t so that 192t is an odd integer (so that cos(96\pi t)= 0) and 8/2= 4 values of t so that 8t is an odd integer (and cos(4\pi t)= 0). That gives a total of 90 values of t for which cos(96\pi t)cos(4\pi t)= 0.

 

[zorro]

Even I have no idea what you did . Anyway I was wrong earlier.
Why did you divide 192 and 8 by 2? What do we get by doing that?

 

[zorro]

Somebody help me out!

 

[HallsofIvy]

You are looking for t such that 192t is an odd integer. 192t= 1 if t= 1/192, of course, but also, 192t= 3 if t= 3/192, 192t= 5 if t= 5/192, etc. We can do that for every odd integer up to 192- and 192/2 of the integers less than 192 are odd.

 

[zorro]

ohh..its that way!. Sometimes simple things become very complex.
I understood it now. Thank you